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3 years ago

If a lamp has a resistance of 50 ohms and is operated by a p.d. of 10V, find the current flowing through it

Physics

1 answer:

NemiM [27]3 years ago

4 0

V=IR therefore I=V/R=10/50=0.2A therefore the current is 0.2 A

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Answer:

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3 years ago

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asambeis [7]

Answer:

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Explanation:

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3 years ago

What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface

kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman $= \frac{95.08}{100 } = 0.9508$

The density of the woman is calculate as;

$Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water$

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0

3 years ago

Select the situation for which the torque is the smallest.

alina1380 [7]

Answer:

e. The torque is the same for all cases.

Explanation:

The formula for torque is:

τ = Fr

where,

τ = Torque

F = Force = Weight (in this case) = mg

r = perpendicular distance between force an axis of rotation

Therefore,

τ = mgr

Here,

m = 200 kg

r = 2.5 m

Therefore,

τ = (200 kg)(9.8 m/s²)(2.5 m)

τ = 4900 N.m

Here,

m = 20 kg

r = 25 m

Therefore,

τ = (20 kg)(9.8 m/s²)(25 m)

τ = 4900 N.m

Here,

m = 8 kg

r = 62.5 m

Therefore,

τ = (8 kg)(9.8 m/s²)(62.5 m)

τ = 4900 N.m

Hence, the correct answer will be:

e. The torque is the same for all cases.

6 0

3 years ago

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

Natali [406]

centripetal acceleration is given by formula

$a_c = \omega^2*R$

given that

$a_c = 34.1 m/s^2$

$R = 5.91 m$

now we have

$\omega^2 R = 34.1$

$\omega^2 * 5.91 = 34.1$

$\omega^2 = 5.77$

$\omega = 2.4 rad/s$

so the ratationa frequency is given by

$\omega = 2 \pi f$

$2.4 = 2 \pi f$

$f = \frac{2.4}{2\pi}$

$f = 0.38 Hz$

7 0

3 years ago